Question: Let $h$ be a vector-valued function defined by $h(t)=(3\cos(t),-\sqrt{4t})$. Find $h'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $(-3\cos(t),-2t)$ (Choice B) B $\left(3\sin(t),-\dfrac2{\sqrt t}\right)$ (Choice C) C $\left(-3\sin(t),-\dfrac1{\sqrt t}\right)$ (Choice D) D $-3\sin(t)+\dfrac2{\sqrt t}$
Answer: $h$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $h(t)=(3\cos(t),-\sqrt{4t})$. Let's differentiate the first expression: $\dfrac{d}{dt}[3\cos(t)]=-3\sin(t)$ Let's differentiate the second expression: $\begin{aligned}\dfrac{d}{dt}(-\sqrt{4t})&=-\dfrac12\cdot\dfrac2{\sqrt t} \\\\&=-\dfrac1{\sqrt t}\end{aligned}$ Now let's put everything together: $\begin{aligned} h'(t)&=\left(\dfrac{d}{dt}[3\cos(t)],\dfrac{d}{dt}(-\sqrt{4t})\right) \\\\ &=\left(-3\sin(t),-\dfrac1{\sqrt t}\right) \end{aligned}$ In conclusion, $h'(t)=\left(-3\sin(t),-\dfrac1{\sqrt t}\right)$.